# Finite Math Question

Two people are chosen at random from a group of eight people consisting of 4

married couples. What is the probability that the two people chosen are not married to each other?

A. (8 choose 6) / (8 choose 2)

B. (8 choose 2) / ( 8 choose 4)

C. 1/7

D. 1/4

E. 6/7

### 2 個解答

• 6 年前
最愛解答

Method 1

No matter who is chosen as the first one, in the remaining 7 people, 1 of them is married to the first chosen person and 6 of them is not married to the first chosen person.

Therefore, the required probability

= Pr(the two people chosen are not married to each other)

= 6/7 (E)

Method 2

Denote:

Couple 1 = {Husband 1, Wife 1} = {H1, W1}

Couple 2 = {Husband 2, Wife 2} = {H2, W2}

Couple 3 = {Husband 3, Wife 3} = {H3, W3}

Couple 4 = {Husband 4, Wife 4} = {H4, W4}

If two people chosen are not married to each other, they must be from different couples, then first choose 2 couples out of 4, i.e., (4 choose 2).

For each of the two chosen couples, choose 1 member from either husband or wife, i.e., (2 choose 1).

The total number of ways to choose 2 out of 8 is (8 choose 2).

Therefore, the required probability

= Pr(the two people chosen are not married to each other)

= (4 choose 2) (2 choose 1) (2 choose 1) / (8 choose 2)

= (6 × 2 × 2) / 28

= 6/7 (E)

2014-11-06 08:33:25 補充：

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• kc
Lv 7
6 年前

我認為功課還是自己做的好.

希望不要有人為了好心幫人, 而淪為別人的鎗手.